上海大学大一高数答案第1章
《高等数学教程》第一章 习题答案
习题1-1 (A)
1.(1)(-∞, 1) ⋃(1, 2) ⋃(2, +∞) (2)[-1, 0) ⋃(0, 1]
(3)(-∞, -1) ⋃(-1, 1) ⋃(1, +∞) (4)x ≠k π且x ≠k π+π
2
(k =0, ±1, ±2, ) (5)(2k π+π
5π3
, 2k π+
3
) (k =0, ±1, ±2, )
(6)[-1, 3] 2. , 6+1
, 6+(x 29
20+h ) 3. 122
,
2, 22
, 0 5.(1)奇函数 (2)非奇非偶函数 (3)偶函数 (4)奇函数 (5)奇函数
(6)当f (x ) 为奇函数或偶函数时,该函数为偶函数;
当f (x ) 为非奇非偶函数时,该函数为非奇非偶函数. (7)偶函数 (8)奇函数
6.(1)是周期函数,T =2π (2)是周期函数,T =4 (3)是周期函数,T =4 (4)不是周期函数
7.(1)y =
-dx +b cx -a (2)y =13arcsin x
2
(3)y =e x -1-2 (4)y =log x
21-x
(5)e x -e -x
y =2
1
8.(1)y =u , u =a -x 2 (2)y =e u , u =x 2 (3)y =lg u , u =cos (4)y =u 2, u =tgv , v =6x (5)y =arctgu , u =cos v , v =e w , w =- (6)y =u 2, u =ln v , v =ln w , w =x 2
9.(1)[-1, 1] (2) [2k π, (2k +1) π] (3)[-a , 1-a ]
1 2x
k ∈z
(4)若0122
,则D =Ф. 10. ϕ[ϕ(x )]=x 4,ψ[ψ(x )]=22x
,ϕ[ψ(x )]=22x ,ψ[ϕ(x )]=2x 2
. 11. a =4, b =-1
⎧1, x
0⎨0, x =0
,g [f (x )]=⎪x
⎪⎨1, x =1
⎩
-1, x >0⎪⎪⎩e -1, x >1
13. V =πh [r 2-(h 2
) 2], (0
14. V =r 324π
2
(2π-α) 24πα-α2
, 0
πr 2h 3
3[(h -r ) 2-r 2]
, (2r , +∞)
⎧900≤x ≤10016.(1)p =⎪
,
⎨90-(x -100) ⋅0. 01, 100
⎪⎩
75, x ≥1600⎧30x ,
0≤x ≤100 (2) p =(p -60) x =⎪
⎨31x -0. 01x 2, 100
⎪⎩
15x , x ≥1600 (3)p =21000(元)
2
习题1-1 (B)
1. f (x ) 为偶函数. 2. f (x ) =x 2-2, f (x -1) =x 2+
1
x
x
2-4 3. f [g (x )]=⎧⎨0, x
,g [f (x )]=⎧⎩x 2
, x ≥0
⎨0, x
⎩x 2
, x ≥0
4. 3+2x 21+x 2
8. f (x ) =⎧⎨1-e -x , -1
⎩-1,
x ≤-1
9. g (x ) =ln(1-x ) , (-∞, 0]
10. 奇函数,偶函数,偶函数,偶函数. 12. f (2005) =1
习题1-2 (A)
1.(1)
1
2n +1
, 0 (2)(-1) n +1
1
n +1
, 0 (3)n
+2
, 1 (4)(n +1) ⋅(-1) n +1n , 没有极限 (5)
12n +1(n +1) 2+(n +1) 2+ +
(n +1) 2, 12
(n +1)(n +2)
(6)(-1) 2
, 没有极限.
2.(1)17; (2)24; (3)[3
ε
]
3.0, [1ε
]
习题1-3 (A)
3. δ=0. 0002 4. Z ≥
3
f (x ) =1 6. lim f (x ) =lim f (x ) =1, lim x →0
x →0-
x →0+
lim x →0-
ϕ(x ) =-1, x lim →0+
ϕ(x ) =1, lim x →0
ϕ(x ) 不存在.
习题1-4 (A)
3.(1)0; (2)0; (3)0
4. x lim →-1y =0; lim x →1
y =∞ 习题1-4 (B)
3. y =x cos x 在(-∞, +∞) 上无界,但当x →+∞时,此函数不是无穷大. 5. 当a =0, b =1时,f (x ) 是无穷小量; 当a ≠0, b 为任意实数时,f (x ) 是无穷大量.
习题1-5 (A)
1.(1)0; (2)1; (3)1; (4)3
10
; (5)
a -13a
2
; (6)3x 2
; (7)43; (8)-1. 2.(1)-34; (2)0; (3)∞; (4)-1
4;
(5)220⋅330
15
50; (6) -4.
⎧1,
0
3.(1)⎪
⎨0,
a =1; (2)3; (3)41
⎪⎩
-1, a >1
3; (4)-2
4.(1)10; (2)
mn (n -m ) 2; (3)m
n ; (4)0; (5)0; (6)12; (7)34; (8)1
2
.
4
习题1-5 (B)
1.(1)2; (2)-; (3)-
12
12a (3-1) ; (4) 562
⎧0, k >2
3⎪
(5); (6)⎨1, k =2; (7)2; (8)0 .
22. α=1, β=-1 3. a =9 4. a =1, b =-1 5. 不一定.
1.(1)2; (6)π2
; 2.(1)e -1; (4)e -2; 1.(1)12
; (5)0; 2.(4)3; 1. 当x →0时,
⎪⎩
∞, k
习题1-6 (A)
(2)3; (3)12
; (4)-1; (7)1; (8)2; (9)1; (2)e 2; (3)e -2; (5)e -1; (6)e 2.
习题1-6 (B)
(2)2
π
; (3)1; (4)0;
(6)1; (7)0; (8)e -1. (5)
1+2
. 习题1-7 (A)
x 4-x 3比x 2+x 3为高阶无穷小.
(5)cos a ; (10)x . 5
2. (1)同阶,但不是等价; (2)同阶,且为等价. 3. α=12
4. α=m
6.(1)3⎧0, m
2; (2)⎪
⎨1, m =n ; (3)1;
⎪⎩
∞, m >n
2 (4)1; (5)a ; (6)12b 4
.
习题1-7 (B)
1.(1)2; (2)e ; (3)1322
; (4)0; (5)1; (6)-14
; (7)∞; (8)1. 5. p (x ) =2x 3+x 2+3x . 6. A ln a .
习题1-8 (A)
1. a =1
2. f (x ) 在x =0处连续
3.(1)x =1为可去间断点,补充f (1) =-2
x =2为第二类间断点
(2)x =0和x =k π+π为可去间断点,补充f (0) =1, f (π
2
k π+2
) =0;x =k π(k ≠0) 为第二类间断点.
(3)x =1为第一类间断点 (4)x =0为第二类间断点.
6
4.(1)x =1为可去间断点,补充f (1) =23;
(2)x =0为可去间断点,补充f (0) =1
2
;
(3)x =1为可去间断点,补充f (1) =-π
2;x =0为第二类间断点;
(4)x =2为可去间断点,补充f (2) =1
4
;x =0为第一类间断点;
x =-2为第二类间断点. (5)x =0为第一类间断点; (6)x =a 为第一类间断点; (7)x =1为第一类间断点; (8)x =-1为第二类间断点.
习题1-8 (B)
1. x =±1为第一类间断点. 2. a =0, b =1 3. a =52
4. a =2n π-
π
2
(n =0, ±1, ±2, )
5. a =-π, b =0
6. (1)当a =0, b ≠1时,有无穷间断点x =0; (2)当a ≠1, b =e 时,有无穷间断点x =1.
习题1-9 (A)
1. 连续区间为:(-∞, -3), (-3, 2), (2, +∞)
l i m f (x ) =1
,lim f (x ) =-8x →
02x →-3
5
,lim x →2f (x ) =∞.
7
2. 连续区间为:(-∞, 0), (0, +∞) .
3. (1) -1; (2) 1; (3) h ; (4) -1; (5) -
2
; (6) -2; (7) 1; (8) 1; 2
(9) ab ; (10) e 5; (11) -1; (12) 2. 4. a =1 5. a =1
1. (1)x =0为第一类间断点; (3)x =0为第一类间断点; (5)无间断点. 2. a =0, b =1
3. (1)e -1
; (2)e -
12
; (5)0; (6)-2; 4.
1
2
一. 1. D 2. D 6. D 7. D 二.1. f (-x ) =⎧⎪⎨x 2-x , x
⎪2x ≥0
⎩x , 2. arcsin(1-x 2) , [-2, 2] 3. -1
习题1-9 (B)
(2)x =-1为第一类间断点; (4)x =±1为第一类间断点; (3)e cot a ; (4)0;
(7)1π2
2; (8)8
.
总复习题一
3. D 4. B 5. C 8. C 9. D 10. D
8
4. 必要,充分 5. 必要,充分 6. 充分必要 7.
12
8. a =b 9.
65
10. 第二类,第一类 三. 1. ϕ(x ) =
x +1x -1 2. α=-20042005, β=1
2005
3. lim n →∞x n =1 4. 4 5. e 4 6. -50 7.
1
2
ln a 8. 当α≤0时,f (x ) 在x =0处不连续;
当α>0, β=-1时,f (x ) 在x =0处不连续; 当α>0, β≠-1时,f (x ) 在x =0处不连续. 9. -28
部分习题选解 习题1-2 (B)
1. 根据数列极限的定义证明:
(1)lim n →∞
a =1(a >0时)
证明:(ⅰ) ∀ε>0
当a >1时,令a =1+h n (h n >0) ∴a =(1+h n n (n -1) n ) =1+nh n +2
h 2n
n + +h n >nh n ∴0
n
a
ε
9
∴取N =[a
ε
]+1,当n >N 时,
有a -1=h a
n
n
(ⅱ) 当a =1时,显然成立. (ⅲ) 当0
1
a
>1 ∴lim b 1n →∞=lim n →∞
n
a
=1
∴lim n
n →∞
a =1 综合(ⅰ) ,(ⅱ) ,(ⅲ) ,∴当a >0时,有lim n →∞
a =1. 习题1-6 (B)
3. 设x 0, y 0>0,x x n +y n
n +1=x n y n ,y n +1=
2
. 证明:lim n →∞x n =lim n →∞
y n 证明: x n y +y n
n ≤
x n 2
∴0≤x n +1≤y n +1(n =0, 1, 2, )
∴x
n +1=x n y n ≥x n x n =x n
y =x n +y n y n +y (n =0, 1, 2, ) n +1
2≤n
2
=y n
由此可知数列{x n }单调增加,数列{y n }单调减少, 又x 0≤x 1≤ ≤x n ≤x n +1≤y n +1≤y n ≤ ≤y 1≤y 0 ∴{x n }与{y n }都是有界的.
由“单调有界数列必有极限”准则, ∴{x n },{y n }都收敛.
设lim
n →∞x n =a , lim n →∞
y n =b 10
由y n +1x n +y n =y n ,∴lim n →∞2=lim x n +y n n →∞2 ∴b =a +b ⇒a =b 2
x n =lim y n . 即lim n →∞n →∞
习题1-10 (B)
3. 设函数f (x ) 在[0, 1]上非负连续,且f (0) =f (1) =0, 试证:对∀l ∈(0, 1) ,必存在一点x 0∈[0, 1-l ],使f (x 0) =f (x 0+l ) . 证明:令F (x ) =f (x ) -f (x +l ) , ∀l ∈(0, 1) f (x ) 在[0, 1]上连续,f (x +l ) 在[-l , 1-l ]上连续, ∴F (x ) 在[0, 1-l ]上连续.
F (0) =f (0) -f (l ) =-f (l ) ≤0 又 ( f (x ) ≥0) F (1-l ) =f (1-l ) -f (1) =f (1-l ) ≥0
∴F (0) ⋅F (1-l ) ≤0
(ⅰ) 若F (0) =0,取x 0=0,即f (0) =f (l ) (ⅱ) 若F (1-l ) =0,取x 0=1-l ,即f (1-l ) =f (1) (ⅲ) F (0) ≠0, F (1-l ≠0) ∴F (0) ⋅F (1-l )
使F (x 0) =0, 即f (x 0) =f (x 0+l ) . 综合(ⅰ) ,(ⅱ) ,(ⅲ) ,对∀l ∈(0, 1) ,必存在一点x 0∈[0, 1-l ],使f (x 0) =f (x 0+l ) .
总复习题一
三.11. 设f (x ) 在[a , b ]上连续,且f (x ) 在[a , b ]上无零点. 证明f (x ) 在[a , b ]上不变号. 11
证明:(反证法) 假设f (x ) 在[a , b ]变号, 即∃x 1, x 2∈[a , b ],使f (x 1) >0, f (x 2)
12